Question

Find the points of intersection of the polar curves

r=1 and r²=2 sin (2 θ)
Find the area of the region that lies inside r²=2 sin (2 θ)

Answer 1

The points of intersection of the polar curves r=1 and r²=2 sin(2θ) are (1, π/12) and (1, 5π/12). The area of the region that lies inside r²=2 sin(2θ) is √3/4 square units.

Step-by-step explanation:

The points of intersection of the polar curves r=1 and r²=2 sin(2θ) can be found by setting the two equations equal to each other and solving for θ:

r² = 2 sin(2θ)

1 = 2 sin(2θ)

sin(2θ) = 1/2

From the unit circle, we know that sin(π/6) = 1/2 and sin(5π/6) = 1/2. Therefore, 2θ can equal π/6 or 5π/6. Solving for θ:

2θ = π/6 or 2θ = 5π/6

θ = π/12 or θ = 5π/12

So the points of intersection are (r, θ) = (1, π/12) and (1, 5π/12).

The area of the region that lies inside r²=2 sin(2θ) can be found by integrating the equation r²=2 sin(2θ)^:

A = ∫[π/12, 5π/12] 1/2 sin(2θ) dθ

Using the double-angle identity sin(2θ) = 2sin(θ)cos(θ), the integral becomes:

A = ∫[π/12, 5π/12] 1/2 (2sin(θ)cos(θ)) dθ

Simplifying the equation, we get:

A = ∫[π/12, 5π/12] sin(θ)cos(θ) dθ

Using the half-angle identity sin(θ)cos(θ) = 1/2 sin(2θ), the integral becomes:

A = ∫[π/12, 5π/12] 1/2 sin(2θ) dθ

Evaluating the integral:

A = (-1/4)(cos(5π/6) - cos(π/6))

A = (-1/4)(-√3/2 - √3/2)

A = (-1/4)(-√3)

A = √3/4

Therefore, the area of the region that lies inside r²=2 sin(2θ) is √3/4 square units.